Question
If \[ A = \begin{bmatrix} 1 & 0 & -3 \\ 2 & 1 & 3 \\ 0 & 1 & 1 \end{bmatrix} \] verify that \[ A^2 + A = A(A + I) \] where \(I\) is the identity matrix.
Solution
Step 1: Compute \(A^2\)
\[ A^2 = \begin{bmatrix} 1 & -3 & -6 \\ 4 & 4 & 0 \\ 2 & 2 & 4 \end{bmatrix} \]Step 2: Find LHS
\[ A^2 + A = \begin{bmatrix} 1 & -3 & -6 \\ 4 & 4 & 0 \\ 2 & 2 & 4 \end{bmatrix} + \begin{bmatrix} 1 & 0 & -3 \\ 2 & 1 & 3 \\ 0 & 1 & 1 \end{bmatrix} = \begin{bmatrix} 2 & -3 & -9 \\ 6 & 5 & 3 \\ 2 & 3 & 5 \end{bmatrix} \]Step 3: Compute \(A + I\)
\[ A + I = \begin{bmatrix} 2 & 0 & -3 \\ 2 & 2 & 3 \\ 0 & 1 & 2 \end{bmatrix} \]Step 4: Find RHS
\[ A(A + I) = \begin{bmatrix} 2 & -3 & -9 \\ 6 & 5 & 3 \\ 2 & 3 & 5 \end{bmatrix} \]Final Result
\[
A^2 + A = A(A + I)
\]
Hence verified.